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  <div class="question_difficulty">
   难度：Easy
  </div>
  <div>
   <h1 class="question_title">
    789. Kth Largest Element in a Stream
   </h1>
   <p>
    Design a class to find&nbsp;the
    <strong>
     k
    </strong>
    th largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.
   </p>
   <p>
    Your&nbsp;
    <code>
     KthLargest
    </code>
    &nbsp;class will have a constructor which accepts an integer
    <code>
     k
    </code>
    and an integer array
    <code>
     nums
    </code>
    , which contains initial elements from&nbsp;the stream. For each call to the method
    <code>
     KthLargest.add
    </code>
    , return the element representing the kth largest element in the stream.
   </p>
   <p>
    <strong>
     Example:
    </strong>
   </p>
   <pre>
int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);&nbsp; &nbsp;// returns 4
kthLargest.add(5);&nbsp; &nbsp;// returns 5
kthLargest.add(10);&nbsp; // returns 5
kthLargest.add(9);&nbsp; &nbsp;// returns 8
kthLargest.add(4);&nbsp; &nbsp;// returns 8
</pre>
   <p>
    <strong>
     Note:
    </strong>
    <br>
    You may assume that&nbsp;
    <code>
     nums
    </code>
    ' length&nbsp;&ge;&nbsp;
    <code>
     k-1
    </code>
    &nbsp;and
    <code>
     k
    </code>
    &ge;&nbsp;1.
   </p>
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   <h1 class="question_title">
    789. 数据流中的第K大元素
   </h1>
   <p>
    设计一个找到数据流中第K大元素的类（class）。注意是排序后的第K大元素，不是第K个不同的元素。
   </p>
   <p>
    你的&nbsp;
    <code>
     KthLargest
    </code>
    &nbsp;类需要一个同时接收整数&nbsp;
    <code>
     k
    </code>
    和整数数组
    <code>
     nums
    </code>
    &nbsp;的构造器，它包含数据流中的初始元素。每次调用&nbsp;
    <code>
     KthLargest.add
    </code>
    ，返回当前数据流中第K大的元素。
   </p>
   <p>
    <strong>
     示例:
    </strong>
   </p>
   <pre>
int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);&nbsp; &nbsp;// returns 4
kthLargest.add(5);&nbsp; &nbsp;// returns 5
kthLargest.add(10);&nbsp; // returns 5
kthLargest.add(9);&nbsp; &nbsp;// returns 8
kthLargest.add(4);&nbsp; &nbsp;// returns 8
</pre>
   <p>
    <strong>
     说明:
    </strong>
    <br>
    你可以假设&nbsp;
    <code>
     nums
    </code>
    &nbsp;的长度&ge;&nbsp;
    <code>
     k-1
    </code>
    &nbsp;且
    <code>
     k
    </code>
    &ge;&nbsp;1。
   </p>
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